Question 890645: The statistics-conscious store manager at Marketview records the number of customers who walk though the door each day. Years of records show the mean number of customers per day to be 586 with a standard deviation of 186. Assume the number of customers is normally distributed.
a) What is the probability that on any given day the number of customers exceeds 1000?
b) If 20 days are randomly selected, what is the probability that the mean of this sample is less than 500?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The statistics-conscious store manager at Marketview records the number of customers who walk though the door each day. Years of records show the mean number of customers per day to be 586 with a standard deviation of 186. Assume the number of customers is normally distributed.
a) What is the probability that on any given day the number of customers exceeds 1000?
z(1000) = (1000-586)/186 = 2.226
P(x > 1000) = P(z > 2.226) = normalcdf(2.226,100) = 0.0130
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b) If 20 days are randomly selected, what is the probability that the mean of this sample is less than 500?
t(500) = (500-586)/[186/sqrt(20)] = -2.068
P(x-bar < 500) = P(t < -2.078) = 0.0193
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Cheers,
Stan H.
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